Us Solutions Manual To Accompany Elements Of Physical Chemistry 7e
Us Solutions Manual To Accompany Elements Of Physical Chemistry 7e
7th Edition
ISBN: 9780198802259
Author: ATKINS
Publisher: Oxford University Press
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Chapter 4, Problem 4.21P

(a)

Interpretation Introduction

Interpretation:

The overall mole fraction of phenol in the given mixture has to be determined.

Concept introduction:

A mole fraction of a molecule in a mixture is the ratio of number of moles of particular molecule to the sum of number of moles of all molecules in the mixture.  Equation for mole fraction of a molecule in a mixture of three molecules (A, B and C) is,

moleculefractionofA,(χA=numbersof moles ofmolecule A(nA)total number of moles(nA+nB+nC

The lever rule is mainly used to determine compositions of phases and the relative proportions of phases to each other in Binary diagrams of compounds and using the lever rule we can determine quantitatively the relative composition of a mixture in a two phase region in a phase diagram.

According to lever rule, we can write,

n=n'+n''

Where,

n': Total amount of molecule in one

n'': Total amount of molecule in other phase

The total amount of A in the sample is nxA where xA it is the overall mole fraction of A in the sample.  The overall amount of A is also the sum of its amounts in the two phases, where it has the mole fractions xA'andxA'' respectively.

Thus,

nxA=n'xA'+n''xA''n'(xA'xA)=n''(xAxA'')

(b)

Interpretation Introduction

Interpretation:

The relative amounts of two phases in the given mixture has to be determined using lever rule.

Concept introduction:

A mole fraction of a molecule in a mixture is the ratio of number of moles of particular molecule to the sum of number of moles of all molecules in the mixture.  Equation for mole fraction of a molecule in a mixture of three molecules (A, B and C) is,

  χA=numbersof moles ofmolecule A(nA)total number of moles(nA+nB+nC

The lever rule is mainly used to determine compositions of phases and the relative proportions of phases to each other in Binary diagrams of compounds and using the lever rule we can determine quantitatively the relative composition of a mixture in a two phase region in a phase diagram.

According to lever rule, we can write,

n=n'+n''

Where,

n': Total amount of molecule in one phase

n'': Total amount of molecule in other phase

The total amount of A in the sample is nxA where xA it is the overall mole fraction of A in the sample.  The overall amount of A is also the sum of its amounts in the two phases, where it has the mole fractions xA'andxA'' respectively.

Thus,

  nxA=n'xA'+n''xA''n'(xA'xA)=n''(xAxA'')

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Chapter 4 Solutions

Us Solutions Manual To Accompany Elements Of Physical Chemistry 7e

Ch. 4 - Prob. 4D.1STCh. 4 - Prob. 4D.2STCh. 4 - Prob. 4D.3STCh. 4 - Prob. 4D.4STCh. 4 - Prob. 4E.1STCh. 4 - Prob. 4E.2STCh. 4 - Prob. 4F.1STCh. 4 - Prob. 4A.1ECh. 4 - Prob. 4A.2ECh. 4 - Prob. 4A.3ECh. 4 - Prob. 4A.4ECh. 4 - Prob. 4A.5ECh. 4 - Prob. 4A.6ECh. 4 - Prob. 4B.1ECh. 4 - Prob. 4B.2ECh. 4 - Prob. 4B.3ECh. 4 - Prob. 4B.4ECh. 4 - Prob. 4B.5ECh. 4 - Prob. 4B.6ECh. 4 - Prob. 4B.7ECh. 4 - Prob. 4B.8ECh. 4 - Prob. 4C.1ECh. 4 - Prob. 4C.2ECh. 4 - Prob. 4C.3ECh. 4 - Prob. 4C.4ECh. 4 - Prob. 4C.5ECh. 4 - Prob. 4C.6ECh. 4 - Prob. 4C.7ECh. 4 - Prob. 4D.1ECh. 4 - Prob. 4D.2ECh. 4 - Prob. 4D.3ECh. 4 - Prob. 4D.4ECh. 4 - Prob. 4D.5ECh. 4 - Prob. 4D.6ECh. 4 - Prob. 4D.7ECh. 4 - Prob. 4D.8ECh. 4 - Prob. 4D.9ECh. 4 - Prob. 4D.10ECh. 4 - Prob. 4D.11ECh. 4 - Prob. 4D.12ECh. 4 - Prob. 4D.13ECh. 4 - Prob. 4E.1ECh. 4 - Prob. 4E.2ECh. 4 - Prob. 4E.3ECh. 4 - Prob. 4E.4ECh. 4 - Prob. 4F.1ECh. 4 - Prob. 4F.2ECh. 4 - Prob. 4F.3ECh. 4 - Prob. 4.1DQCh. 4 - Prob. 4.2DQCh. 4 - Prob. 4.3DQCh. 4 - Prob. 4.4DQCh. 4 - Prob. 4.5DQCh. 4 - Prob. 4.6DQCh. 4 - Prob. 4.8DQCh. 4 - Prob. 4.9DQCh. 4 - Prob. 4.10DQCh. 4 - Prob. 4.11DQCh. 4 - Prob. 4.12DQCh. 4 - Prob. 4.13DQCh. 4 - Prob. 4.14DQCh. 4 - Prob. 4.15DQCh. 4 - Prob. 4.1PCh. 4 - Prob. 4.2PCh. 4 - Prob. 4.3PCh. 4 - Prob. 4.4PCh. 4 - Prob. 4.5PCh. 4 - Prob. 4.6PCh. 4 - Prob. 4.7PCh. 4 - Prob. 4.8PCh. 4 - Prob. 4.9PCh. 4 - Prob. 4.10PCh. 4 - Prob. 4.11PCh. 4 - Prob. 4.12PCh. 4 - Prob. 4.13PCh. 4 - Prob. 4.14PCh. 4 - Prob. 4.15PCh. 4 - Prob. 4.16PCh. 4 - Prob. 4.17PCh. 4 - Prob. 4.18PCh. 4 - Prob. 4.19PCh. 4 - Prob. 4.20PCh. 4 - Prob. 4.21PCh. 4 - Prob. 4.22PCh. 4 - Prob. 4.23PCh. 4 - Prob. 4.24PCh. 4 - Prob. 4.25PCh. 4 - Prob. 4.26PCh. 4 - Prob. 4.27PCh. 4 - Prob. 4.28PCh. 4 - Prob. 4.29PCh. 4 - Prob. 4.30PCh. 4 - Prob. 4.1PRCh. 4 - Prob. 4.2PRCh. 4 - Prob. 4.3PRCh. 4 - Prob. 4.4PRCh. 4 - Prob. 4.5PRCh. 4 - Prob. 4.6PR
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Solutions: Crash Course Chemistry #27; Author: Crash Course;https://www.youtube.com/watch?v=9h2f1Bjr0p4;License: Standard YouTube License, CC-BY